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Novosadov [1.4K]
3 years ago
7

Lake Preston's latitude is about 44.4 degrees north of the equator. Which global wind belt are

Physics
1 answer:
Serggg [28]3 years ago
5 0

Answer:

Idk

Explanation:

Idk

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N
9966 [12]

Answer: de?

Explanation:

7 0
3 years ago
Sedimentary layers that are deposited on an angle are called?
Akimi4 [234]

Sedimentary rocks are deposited in layers as strata, forming a structure called bedding.

<span>Bedding planes are surfaces that separate one layer from another. Bedding planes can also form when the upper part of a sediment layer is eroded away before the next episode of deposition. Strata separated by a bedding plane may have different grain sizes, grain compositions, or colors. Sometimes these other traits are better indicators of stratification as bedding planes may be very subtle.</span>

8 0
3 years ago
The radiation per unit area from the Sun reaching the earth is 1400 W/m2 , approximately the amount of radiative power per unit
natka813 [3]

Answer:

83.2 W/m^2

Explanation:

The radiation per unit area of a star is directly proportional to the power emitted, which is given by Stefan-Boltzmann law:

P=\sigma A T^4

where

\sigma is the Stefan-Boltzmann constant

A is the surface area

T is the surface temperature

So, we see that the radiation per unit area is proportional to the fourth power of the temperature:

I \propto T^4

So in our problem we can write:

I_1 : T_1^4 = I_2 : T_2^4

where

I_1 = 1400 W/m^2 is the power per unit area of the present sun

T_1 = 5800 K is the temperature of the sun

I_2 is the power per unit area of sun X

T_2 = 2864 K is the temperature of sun X

Solving for I2, we find

I_2 = \frac{I_1 T_2^4}{T_1^4}=\frac{(1400 W/m^2)(2864 K)^4}{(5800 K)^4}=83.2 W/m^2

6 0
3 years ago
A hair dryer is basically a duct of constant diameter in which a few layers of electric resistors are placed. A small fan pulls
nekit [7.7K]

Given:

density of air at inlet, \rho_{a} = 1.20 kg/m_{3}

density of air at inlet, \rho_{b} = 1.05 kg/m_{3}

Solution:

Now,

\dot{m} = \dot{m_{a}} = \dot{m_{b}}

\rho_{a} A v_{a} = \rho _{b} Av_{b}                        (1)

where

A = Area of cross section

v_{a} = velocity of air at inlet

v_{b} = velocity of air at outlet

Now, using eqn (1), we get:

\frac{v_{b}}{v_{a}} = \frac{\rho_{a}}{\rho_{b}}

\frac{v_{b}}{v_{a}} = \frac{1.20}{1.05} = 1.14

% increase in velocity = 1.14\times 100 =114%

which is 14% more

Therefore % increase in velocity is 14%

5 0
3 years ago
Read 2 more answers
A reconnaissance plane flies 404 km awayfrom its base at 730 m/s, then flies back to its base at 1095 m/s.What is it’s average s
elena-14-01-66 [18.8K]

The average speed of the plane is 875.999 m/s.

Average speed can be defined as the ratio of total distanced travelled by the object to that of total time taken to cover the distance.

Mathematically, Average speed = Av = \frac{Total Distance}{Total Time}

According to the question,

Speed of the plane away from its base V₁ = 730 m/s

Speed of the plane when it flies back V₂ = 1095 m/s

Plane flies the distance D = 404 km

Total Distance covered by the plane S = 404 * 2 km

(because the distance travelled by the plane when going away from the base and then flying back to the base is same)

Therefore S = 808 km = 808 ˣ 10³ m

Time taken by the plane while flying away from the base T₁ = \frac{D}{V1}

T₁ =  \frac{404000}{730} = 553.425 s

Time taken by the plane while flying back to the base T₂ = \frac{D}{V2}

T₂ =  \frac{404000}{1095} = 368.949s

Total Time T = T₁ + T₂ = 922.375 s

Therefore  Av = \frac{Total Distance}{Total Time}

= \frac{D}{T} m/s

=  \frac{808000}{922.375}  m/s

= 875.999 m/s

The average speed of the plane will be 875.999 m/s.

To know more about Average speed,

brainly.com/question/28641761

#SPJ1

7 0
1 year ago
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